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Suppose we have Heisenberg equation of motion for some observable $A$,

$$ i\hbar\frac{dA}{dt}= -[H,A] $$ since the trace of any finite dimensional commutator structure vanish(not something like $[x,p]=i\hbar$ ),

$$ Tr\left(\frac{dA}{dt}\right)=\frac{i}\hbar~Tr([H,A])=0 $$

so my question is, what does $Tr(\frac{dA}{dt})=0$ indicate?

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  • $\begingroup$ $A$ and $H$ are typically not finite dimensional, the energy spectrum is typically infinite, so you cannot use commutativity of Trace. $\endgroup$ – user7757 Jun 12 '14 at 19:22
  • $\begingroup$ @ramanujan_dirac I am not talking about [x,p]=ih, in many cases people are using heisenberg equation of motion, and applied to finite dimensional system. $\endgroup$ – Lorniper Jun 12 '14 at 19:27
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If the trace of an operator with a complete set of eigenvectors is well defined, then it is equal to the sum of the eigenvalues of the operator.

Time evolution of an operator is a unitary transformation, which leave the eigenvalues of an operator unchanged. This means if $a_i$ is an eigenvalue of an operator $A$

\begin{equation}\frac{\mathrm{d}a_i}{\mathrm{d}t} = 0\end{equation}

and so clearly

\begin{equation}\mathrm{Tr}\left(\frac{\mathrm{d}A}{\mathrm{d}t}\right) = \sum_i\frac{\mathrm{d}a_i}{\mathrm{d}t} = 0\end{equation}

So physically this result is to do with the fact that although the system may evolve with time, the spectrum of allowed results for a measurement of $A$ does not.

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  • $\begingroup$ "Time evolution of an operator is a unitary transformation", I don't think this's always true, right? $\endgroup$ – Lorniper Jun 13 '14 at 5:38
  • $\begingroup$ We can write $A = U^\dagger A_0U$, where $A_0$ is time independent. The solution to the Heisenberg equation for $U$ is then $U = \exp\left(-\imath\frac{Ht}{\hbar}\right)$ which can be shown to be unitary for Hermitian $H$ $\endgroup$ – By Symmetry Jun 13 '14 at 9:05

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